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Question

(D) $(1)$ $p_1$ is the probability that the maximum of the chosen numbers is at least $81$.$p_1 = 1 - P(	ext{maximum} \leq 80) = 1 - (\frac{80}{100})

Vedclass · Accepted Answer

$76.25, 24.50$

Vedclass · Answer

(D) $(1)$ $p_1$ is the probability that the maximum of the chosen numbers is at least $81$.$p_1 = 1 - P(	ext{maximum} \leq 80) = 1 - (\frac{80}{100})^3 = 1 - (\frac{4}{5})^3 = 1 - \frac{64}{125} = \frac{61}{125}$.Thus,$\frac{625}{4} p_1 = \frac{625}{4} 	imes \frac{61}{125} = \frac{5 	imes 61}{4} = \frac{305}{4} = 76.25$.$(2)$ $p_2$ is the probability that the minimum of the chosen numbers is at most $40$.$p_2 = 1 - P(	ext{minimum} \geq 41) = 1 - (\frac{60}{100})^3 = 1 - (\frac{3}{5})^3 = 1 - \frac{27}{125} = \frac{98}{125}$.Thus,$\frac{125}{4} p_2 = \frac{125}{4} 	imes \frac{98}{125} = \frac{98}{4} = 24.50$.

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